Economics Stack Exchange is a question and answer site for professional and academic economists and analysts. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

One measure is to look at standard deviations. If a one standard deviation increase in X leads to a more than a 0.5 (or 1, or ⅓, or whatever) standard deviation increase in Y, then we say that X has a economically significant effect on Y.

Is this standard? And if not, what other measures are there to formalize the notion of "economic significance"?

share|improve this question

Imagine the basic linear regression setup:

$Y_i = \alpha + X_i \beta + \epsilon, \epsilon \sim N(0,\sigma^2)$

Holding everything else fixed, increasing $\sigma$ will increase the standard deviation of $Y$ and therefore, for a fixed relationship between $X$ and $Y$, eventually move the relationship between them to economic insignificance by the standard you present. But the relationship might still be quite economically important. It might be better to instead ask about the standard deviation effects on $\sigma_\hat{Y}$, the standard deviation of the fitted values of $Y$, to ask if the effect is large relative to the total variation that can be explained by the model.

In accounting, they use a concept of immateriality which seems quite similar to economic insignificance. That might be a helpful definition for some problems.
This quote from Signifying Nothing: Reply to Hoover and Siegler by Deirdre N. McCloskey and Stephen T. Ziliak may also be helpful:

The sheer probability statement about one or two standard errors is useless, unless you have judged by what scale a number is large or small for the scientific or policy or personal purpose you have in mind. This applies to the so-called "precision" or "accuracy" of the estimate, too, beloved of Hoover and Siegler — the number we calculate as though that very convenient sampling theory did in fact apply.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.